# Abstract Nonsense

## Necessary and Sufficient Conditions for Isomorphism

Point of post: This post will give a nice result which will give us necessary and sufficient conditions for a map between vector spaces to be an isomorphism. I will cite this post every now and then when I would like to shorten a proof.

Theorem: Let $\mathscr{U}$ and $\mathscr{V}$ be finite-dimensional vector spaces over the field $F$. Then, a linear mapping $f:\mathscr{U}\to\mathscr{V}$ is an isomorphism if and only if there exists bases $\{x_1,\cdots,x_n\}=\mathcal{B}_1$ for $\mathscr{U}$ and $\{y_1,\cdots,y_m\}=\mathcal{B}_2$ for $\mathscr{V}$ such that

$f_{\mid\mathcal{B}_1}:\mathcal{B}_1\to\mathcal{B}_2$

is a bijection.

Proof: Suppose first that there existed bases $\mathcal{B}_1$ and $\mathcal{B}_2$ such that the above is true and assume without loss of generality that $f(x_j)=y_j$. To prove injectivity suppose that $f(x_1)=f(x_2)$. Then, since $\mathcal{B}_1$ is a basis for $\mathscr{U}$ we know that $\displaystyle x_1=\sum_{j=1}^{n}\alpha_j x_j$ and $\displaystyle x_2=\sum_{j=1}^{n}\beta_j x_j$ for some $\alpha_1,\cdots,\alpha_n,\beta_1,\cdots,\beta_n\in F$. Thus, by the linearity of $f$ we have that

$\displaystyle \sum_{j=1}^{n}\alpha_j f(x_j)=f(x_1)=f(x_2)=\sum_{j=1}^{n}\beta_j f(x_j)$

But, this says that

$\displaystyle \sum_{j=1}^{n}\alpha_j y_j=\sum_{j=1}^{n}\beta_j y_j$

and since $\mathcal{B}_2$ is a basis for $\mathscr{V}$ it follows that $\alpha_j=\beta_j,\text{ }j=1,\cdots,n$ and so $x_1=x_2$. Now, to prove surjectivity we merely note that for any $y\in\mathscr{V}$ we have that there is some $\beta_1,\cdots,\beta_m\in\mathcal{B}_2$ such that $\displaystyle y=\sum_{j=1}^{n}\beta_j y_j$. But, it follows that

$\displaystyle y=\sum_{j=1}^{n}\beta_j y_j=\sum_{j=1}^{n}\beta_j f(x_j)=f\left(\sum_{j=1}^{n}\beta_j x_j\right)$

from where surjectivity follows. Thus, $f$ is a bijective linear mapping and thus an isomorphism.

Conversely, it is trivial that if $f$ is an isomorphism that the image under any basis for $\mathscr{U}$ is a basis for $\mathscr{V}$, and since the restriction of a bijection is a bijection the conclusion follows. $\blacksquare$